Post on 03-Feb-2022
Step Functions, Impulse Functions,and the Delta Function
Department of Mathematics and Statistics
September 7, 2012
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 1 / 9
The Unit Step Function
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 2 / 9
The Unit Step Function
Definition
The unit step function or Heaviside function is a function of the form
ua(t) =
{
0, t < a,1, t ≥ a,
where a > 0.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 2 / 9
The Unit Step Function
Definition
The unit step function or Heaviside function is a function of the form
ua(t) =
{
0, t < a,1, t ≥ a,
where a > 0.
y
ta
1
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 2 / 9
The Laplace Transform of the Unit Step Function
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
∫
∞
a
e−st dt
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
∫
∞
a
e−st dt
= limb→∞
−1
se−st
∣
∣
∣
∣
b
a
=
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
∫
∞
a
e−st dt
= limb→∞
−1
se−st
∣
∣
∣
∣
b
a
=1
se−as .
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Difference Between Two Unit Step Functions
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 4 / 9
The Difference Between Two Unit Step Functions
u1(t) − u2(t)
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 4 / 9
The Difference Between Two Unit Step Functions
u1(t) − u2(t)
y
t
1
1 2 3
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 4 / 9
The Difference Between Two Unit Step Functions
u1(t) − u2(t)
y
t
1
1 2 3
L(u1(t) − u2(t)) = 1se−s − 1
se−2s .
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 4 / 9
The Unit Step Function Combined with Other Functions
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
The Unit Step Function Combined with Other Functions
Consider
g(t) =
{
0, t < a,f (t − a), t ≥ a,
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
The Unit Step Function Combined with Other Functions
Consider
g(t) =
{
0, t < a,f (t − a), t ≥ a,
We may write g(t) = ua(t)f (t − a).
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
The Unit Step Function Combined with Other Functions
Consider
g(t) =
{
0, t < a,f (t − a), t ≥ a,
We may write g(t) = ua(t)f (t − a).
t
y
1 2 3 4 5 6 7
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
L(g(t))
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
∫
∞
0e−s(τ+a)f (τ) dτ
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
∫
∞
0e−s(τ+a)f (τ) dτ = e−as
∫
∞
0e−sτ f (τ) dτ
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
∫
∞
0e−s(τ+a)f (τ) dτ = e−as
∫
∞
0e−sτ f (τ) dτ = e−asL(f (t)).
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
The kth Unit Impulse Function
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 7 / 9
The kth Unit Impulse Function
Definition
The kth unit impulse function is
δa,k(t) =
{
k, a < t < a + 1k,
0, 0 ≤ t ≤ a or t ≥ a + 1k,
where a > 0.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 7 / 9
The kth Unit Impulse Function
Definition
The kth unit impulse function is
δa,k(t) =
{
k, a < t < a + 1k,
0, 0 ≤ t ≤ a or t ≥ a + 1k,
where a > 0.
y
t
10
1 2 3
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 7 / 9
The Laplace Transform of the Unit Impulse Function
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt =
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
= e−sa ·1 − e−s/k
s/k.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
= e−sa ·1 − e−s/k
s/k.
By L’Hopital’s rule: limk→∞
L(δa,k(t)) =
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
= e−sa ·1 − e−s/k
s/k.
By L’Hopital’s rule: limk→∞
L(δa,k(t)) = e−sa.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Dirac Delta Function
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 9 / 9
The Dirac Delta Function
Note that limk→∞
L(δ0,k(t)) = 1.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 9 / 9
The Dirac Delta Function
Note that limk→∞
L(δ0,k(t)) = 1.
We let limk→∞
L(δ0,k(t)) = L(δ(t)), where δ(t) is called the Dirac delta
function or delta function.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 9 / 9
The Dirac Delta Function
Note that limk→∞
L(δ0,k(t)) = 1.
We let limk→∞
L(δ0,k(t)) = L(δ(t)), where δ(t) is called the Dirac delta
function or delta function.
So L(δ(t)) = 1 and L−1(1) = δ(t).
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 9 / 9